Quadratic equations
A quadratic equation in x is an equation that can be written in the standard quadratic
form ax2 + bx + c = 0, a ≠ 0. Several methods can be used to solve quadratic equations.
If the quadratic polynomial ax2 + bx + c can be factored over the integers, then the
equation can be solved by factoring and using zero product property.
The zero product property states that if A and B are algebraic expressions, then AB = 0 if
and only if A = 0 or B = 0. This property states that when the product of two factors
equals zero, then at least one of them must be zero. Let’s look at a couple of examples
that apply this property.
Example 1: Solve the quadratic equation: 3x2 + 10x = 8
Our first step is to write the equation in standard quadratic form; i.e., we want one side of
the equal sign to be zero. To achieve a zero on the right side of the equals sign, we must
subtract 8. If we subtract 8 from the right side, we must also subtract 8 from the left side.
3x2 + 10x = 8
3x2 + 10x –8 = 8 – 8
3x2 + 10x –8 = 0
Now we will attempt to factor the left side. Factors of 3x2 are 3x and x. Because the sign
of the constant factor is negative, that indicates the signs in our factors will be different.
(3x -
) (x +
)
or (3x + ) (x - )
Factors of 8 are 1 and 8 or 2 and 4. Let’s substitute these into our factors and see which
produce the middle term, 10x.
(3x - 1) (x + 8) = 3x2 –8 –1x + 24x = 3x2 + 23x – 8 This is not what we want
or (3x + 1) (x - 8) = 3x2 - 8 +1x - 24x = 3x2 - 23x - 8 This is not what we want
(3x - 2) (x + 4) = 3x2 –8 –2x + 12x = 3x2 +10x –8 This is what we want
or (3x + 2) (x - 4) It isn’t necessary to complete this one.
Now we have our factors and can now solve the equation for x by applying the zero
product property. If the product of two expressions is zero at least one of the expression
must be zero. So one of our factors must be zero.
(3x - 2) (x + 4) = 0
3x – 2 = 0
3x – 2 + 2 = 0 + 2
3x = 2
3x/3 = 2/3
x = 2/3
or x + 4 = 0 Now we solve these linear equations.
x+4–4=0-4
x = -4
Let’s check to be certain our solution is correct. We do this by substituting each value for
x into the original equation.
3x2 + 10x = 8
3(2/3)2 + 10(2/3) = 8
3* 4/9 + 20/3 = 8
12/9 + 20/3 = 8
4/3 + 20/3 = 8
24/3 = 8
8=8
Square 2/3 and multiply 10 times 2/3
Multiply 4/9 by 3
Reduce 12/9 to 4/3
Combine numerators since denominators are same.
Reduce 24/3 to lowest terms
3x2 + 10x = 8
3(-4)2 + 10(-4) = 8
3(16) – 40 = 8
48 – 40 = 8
8=8
Both values result in a true statement.
Now we will examine equations of the form ax2 = c. This is also a quadratic equation
where b = 0. We will put this into the form x2 = c/a then take the square root of both
sides of the equation to find a solution.
Example 2: Solve for x: (x – 1)2 = 49
Applying the square root property to this example is straight forward.
(x − 1)2 = 49
(x − 1)2 =
49
Take the square root of both sides of the equation.
x − 1 = ±7
Remember that the square root of a positive value can be positive or negative. You must
consider both possibilities. This gives us two equations to solve:
x–1=7
x –1 + 1 = 7 + 1
x=8
and x – 1 = -7
x – 1 + 1 = -7 + 1
x = -6
Example 3: 4y2 -20 = 0
First we put it into the form Ay2 = C by adding 20 to both sides of the equation.
4y2 - 20 = 0
4y2 - 20 + 20 = 0 + 20
4y2 = 20
Now we divide by 4 to isolate the variable
2
4y
20
=
4
4
2
y =5
Now we take the square root of both sides.
y2 = 5
y= 5
Sometimes our equations do not lend themselves to factoring or to applying the square
root property. We may then attempt to write the equation in the form of a perfect square
so we can factor the equation. This method is known as completing the square. We need
to examine some characteristics of square binomials before we begin.
Consider the following binomial squares and their perfect-square trinomial products.
Square of a Binomial
(x + 6)2
(x – 3)2
Perfect-Square Trinomial
x2 + 12x + 36
X2 –6x + 9
In each perfect-square trinomial the coefficient of x2 is 1, and the constant term of the
perfect-square trinomial is the square of half the coefficient of its x term.
Adding, to a binomial of the form x2 + bx, the constant that makes that binomial a
perfect-square trinomial is called completing the square. For example, to complete the
square of x2 + 8x, add
2
1 
 * 8  = 16
2 
to produce the perfect-square trinomial x2 + 8x + 16.
Completing the square is a powerful method because it can be used to solve any quadratic
equation.
To solve x2 = 6x –13 by completing the square, begin by writing the variable terms on
one side of the equals sign and the constant term on the other side.
x2 = 6x +13
x2 – 6x = 6x +13 –6x
x2 – 6x = 13
Subtract 6x from both sides
2
1 
Now we will complete the square. That is, we will add  * 6  = 9 to both sides of the
2 
equation.
x2 – 6x + 9 = 13 + 9
x2 – 6x + 9 = 22
Now we will factor the left side of the equation
(x-3)2 = 22
Now we will take the square root of both sides of the equation.
(x − 3)2 = 22
(x − 3) = ± 22
Now we solve this linear equation for x.
x − 3 + 3 = ± 22 + 3
x = ± 22 + 3
Example 4: Solve 2 x 2 + 8 x − 15 = 0 by completing the square.
First we put it in the form ax2 + bx = c by adding 15 t both sides of the equation. Then
we will divide each term by 2 so the coefficient of x2 is 1.
2 x 2 + 8 x − 15 = 0
2 x 2 + 8 x − 15 + 15 = 0 + 15
Add 15 to both sides
2
2 x + 8 x = 15
Simplify
2
2x
8 x 15
+
=
Divide each term by 2
2
2
2
15
x 2 + 4x =
2
To complete the square we will add (½ * 4)2 to both sides of the equation.
15
+4
Next we will combine terms on the right side.
2
15 8
x 2 + 4x + 4 = +
2 2
23
x 2 + 4x + 4 =
2
Now we factor the left side of the equation.
(x + 2)2 = 23
2
Now we take the square root of both sides of the equation
x 2 + 4x + 4 =
23
2
We do not leave a fraction under the denominator, so we must rationalize the right hand
side. We multiply by 2/2 to make the denominator a perfect square.
x+2=±
x+2=±
23 2
*
2 2
Now we can extract the root of the denominator.
1
46
2
Now we solve this linear equation for x by subtracting 2 to both sides of the equation.
1
x+2−2=±
46 − 2
2
x+2=±
x=±
1
46 − 2
2
Our last method for solving quadratic equations is to use the quadratic formula. The
quadratic formula is:
If ax2 + bx +c = 0, a ≠ 0, then
− b ± b 2 − 4ac
x=
2a
We can use this to solve any quadratic equation, but generally apply this only when one
of the other solution strategies doesn’t work.
Example: Solve 4x2 – 8x + 1 = 0
a = 4, b = -8, c = 1
x=
− (−8) ± (−8) 2 − 4 * 4 * 1
2*4
8 ± 64 − 16
x=
8
8 ± 48
x=
8
At this point we need to simplify the square root by factoring it as 3 * 16. I selected 3 *
16 because 16 is the largest perfect square that is a factor of 48.
8 ± 3 * 16
8
8±4 3
x=
8
Now we can reduce the fraction by dividing each term by 4.
2± 3
x=
2
x=